# Sakurai page 91: Simple Harmonic Oscillator, trouble understanding

1. May 1, 2013

### omoplata

From page 91 of "Modern Quantum Mechanics, revised edition", by J. J. Sakurai.

Some operators used below are,
$$a = \sqrt{\frac{m \omega}{2 \hbar}} \left(x + \frac{ip}{m \omega} \right)\\ a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}} \left(x - \frac{ip}{m \omega} \right)\\ N = a^{\dagger} a$$

In equation (2.3.16), he finds
$$a \mid n \rangle = \sqrt{n} \mid n - 1 \rangle$$
Next he says it's similarly easy to show equation (2.3.17),
$$a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle$$
I didn't find it to be so easy. Following the same procedure as equation (2.3.16), since from equation (2.3.12a),
$$Na^{\dagger} \mid n \rangle = (n+1) a ^{\dagger} \mid n \rangle$$, I took
$$a^{\dagger} \mid n \rangle = c \mid n + 1 \rangle$$, where $c$ is some constant. Then,
$$\langle n \mid a a^{\dagger} \mid n \rangle = | c |^2$$
To find $a^{\dagger} a$,
$$\begin{eqnarray} a a^{\dagger} & = & \left( \frac{m \omega}{2 \hbar} \right) \left( x + \frac{ip}{m \omega} \right) \left( x - \frac{ip}{m \omega} \right)\\ & = & \frac{H}{\hbar \omega} + \frac{1}{2}\\ \end{eqnarray}$$
Since it is proven in equation (2.3.5) and (2.3.6) that $H = \hbar \omega (N + \frac{1}{2})$, I get,
$$a a^{\dagger} = N + 1$$
Then,
$$\langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N + 1 \mid n \rangle = n + 1$$
Therefore, from above,
$$c = \sqrt{n+1}$$, which gives,
$$a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle$$, which is the desired result.

But is there an easier way to get this?

At the same time,
$$a a^{\dagger} = N^{\dagger}$$
So,
$$\langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N^{\dagger} \mid n \rangle = n^{*}$$
Therefore,
$$n+1 = n^{*}$$
How can this happen? I can't see this happening unless the real component of $n$ tends to be infinity.

Last edited: May 1, 2013
2. May 1, 2013

### Chain

$N^\dagger = (a^\dagger a)^\dagger = a^\dagger (a^\dagger)^\dagger = a^\dagger a \neq a a^\dagger$

Hope that helps :)

3. May 1, 2013

LOL, thanks.

4. May 1, 2013

### Fightfish

The commutation relation
$$\left[\hat{a},\hat{a}^{\dagger}\right] = 1$$
will be very, very useful for the SHM and many other applications involving similar-looking Hamiltonians.